A hoodlum throws a stone vertically downward with an initial speed v0 from the roof of a building, a height h above the ground. (a) How long does it take the stone to reach the ground? (b) What is the speed of the stone at impact? Give your answers in terms of the given variables and g.
To answer these questions, we can use the equations of motion for freely falling objects.
(a) How long does it take the stone to reach the ground?
The motion of the stone is in the vertical direction, and the only force acting on it is gravity. We can use the equation of motion for free fall:
h = v0t + (1/2)gt^2
Where:
h = height of the building
v0 = initial velocity of the stone (downward)
t = time taken to reach the ground
g = acceleration due to gravity (approximately 9.8 m/s^2)
Since the stone is being thrown downward, the initial velocity v0 will be negative. Therefore, the equation becomes:
h = -v0t + (1/2)gt^2
Rearranging the equation and setting h to zero (as that's when the stone reaches the ground), we can solve for t:
(1/2)gt^2 - v0t = 0
t((1/2)gt - v0) = 0
Either t = 0 or (1/2)gt - v0 = 0
Since the stone is not initially at rest (v0 is not zero), the second equation gives the correct solution:
(1/2)gt - v0 = 0
Simplifying the equation:
gt = 2v0
t = (2v0)/g
So, the time taken for the stone to reach the ground is t = (2v0)/g.
(b) What is the speed of the stone at impact?
The speed of the stone at impact can be found by using the equation of motion:
v = v0 + gt
Substituting v0 = -v0 (since it is moving downward) and t = (2v0)/g, we get:
v = -v0 + g((2v0)/g)
Simplifying the equation:
v = -v0 + 2v0
v = v0
So, the speed of the stone at impact is equal to its initial speed, v0.